A student mixes 20% acid solution with a 40% solution produce 20 liters of 28% solution. How many liters of 20% solution acid solution was needed? Please help

12 L of 20% acid solution must be mixed with 8 L of 40% solution to obtain 20 L of 28% solution.Solution:Let us first set up a table. Fill in the table with known valuesThe table is attached belowFrom the attached table, we can set up two equations,Sum of values of two acids = Value of mixture  0.2x + 0.4y=5.6  For convenience, we'll multiply the entire equation by 10, 2 x + 4 y = 56 ------> (1)  Now, Sum of amounts of each acid = Amount of mixture  x + y = 20 ---------> (2) Multiply eqn (2) with 2 to derive the equation into one variable  eqn ( 2) × 2 => 2x + 2y = 40  Subtracting equation eqn (2) from (1),  ( + ) 2 x + 4 y = 56 ( − ) 2 x + 2 y = 40 − − − − − − − − ( = ) 0 + 2y = 16  Thus, 2y = 16y = 8  Substituting y = 8 in eqn (2),x + 8  = 20x = 20 – 8 = 12   So, we have x = 12 and y = 8 We can conclude that 12 L of 20% acid solution must be mixed with 8 L of 40% solution to obtain 20 L of 28% solution.